Laws of Motion Test

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Laws of Motion Test - 65

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Question 1
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A stationary ball weighing $$0.25\ kg$$ acquires a speed of $$10 \ m/s$$ when hit by a hockey stick. The impulse imparted to the ball is:

A
$$2.5\ N s$$

B
$$2.0\ N s$$

C
$$1.5\ N s$$

D
$$0.5 \ N s$$

Solution

Given that the mass of the body is $$0.25 kg$$ is at rest initially. when the body is hit by the stick it gains speed of  $$10\ m/s$$.
means initial speed $$u = 0$$ and after getting hit it's final speed $$ v = 10 m/s$$.
we know that impulse is equal to change in momentum i.e.  $$\text {Impulse } J =m(v-u)$$
so here $$ J = 0.25 ( 10 - 0 )$$  $$kg\ m/s$$ or $$ N $$
hence The impulse imparted to the ball is $$2.5 N s$$
Question 2
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Two ends of a uniform garland of mass $$m$$ and length $$L$$ are hanging vertically as shown. At instant $$t$$, the end Z is released. If $$y$$ is the distance moved by this end in time $$dt$$, change in momentum of an element of length $$dy$$ at a distance $$y$$ from Z is given as:

A
$$\displaystyle \dfrac{m}{L} dy (2 gy)^{\dfrac{1}{2}}$$ (downward)

B
$$\displaystyle \dfrac{m}{L} dy (2 gy)^{\dfrac{1}{2}}$$ (upward)

C
$$\displaystyle \dfrac{-m}{L} dy \, \, dy \sqrt{gy}$$ (upward)

D
$$\displaystyle \dfrac{m}{L} dy\, \, (gy)^{\dfrac{1}{2}}$$ (upward)

Solution

When the end Z goes down by $$y$$,
the velocity of bend $$=v=\sqrt{2g\dfrac{y}{2}}$$
because the bend falls down by $$y/2$$.
Change in momentum of the element
$$=-\dfrac{m}{L}dyv$$ (downward)
$$=\dfrac{m}{L}(gy)^{1/2}dy$$ (upward)
Question 3
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A pendulum was kept horizontal and released. Find the acceleration of the pendulum when it makes an angle $$ \theta $$ with the vertical.

A
$$ g\ \sqrt { 1+3\cos ^{ 2 }{ \theta } } $$

B
$$ g\ \sqrt { 1+3\sin ^{ 2 }{ \theta } } $$

C
$$ g\ \sin { \theta } $$

D
$$ 2g\ \cos { \theta } $$

Solution

$$ \dfrac { { mv }^{ 2 } }{ 2 } $$ = $$ mgl\cos { \theta } $$ or
$$\dfrac { { v }^{ 2 } }{ l } $$ = $$ 2g\ \cos { \theta } $$
and $$ \sqrt { { a }_{ t }^{ 2 }+{ a }_{ r }^{ 2 } } =\sqrt { { \left( g\sin { \theta } \right) }^{ 2 }+{ (2g\quad \cos { \theta } ) }^{ 2 } } $$
= $$ g \ \sqrt { 1+3\cos ^{ 2 }{ \theta } } $$
Question 4
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Directions For Questions

A uniform ladder $$5.0$$ long rests against a frictionless, vertical wall with its lower end $$3.0$$ from the wall. The ladder weighs $$160 N$$. The coefficient of static friction between the foot of the ladder and the ground is $$0.40$$. A man weighing $$740 N$$ climbs slowly up the ladder.

...view full instructions

What is the actual frictional force when the man has climbed $$1.0$$ m along the ladder?

A
$$360 N$$

B
$$171 N$$

C
$$900 N$$

D
$$740 N$$

Solution

At the point on the vertical wall there is normal force along horizontal direction and at the ground there will be normal force in vertical direction and friction force in horizontal direction.
Weight of Ladder is acting at 1.5m from ground contact point and is 160 N (downwards) in vertical direction and Weight of man is acting at (3/5)x1=0.6m from ground contact and is 740 N (downwards) in vertical direction

Net torque about ground is zero,
$$ 740\times 0.6+160\times 1.5-{ N }_{ H }\times 4=0\\ { N }_{ H }=\quad 171\quad N $$

By doing Force Balance along horizontal we get Friction force=Normal Horizontal= 171N

Just to check whether friction can be upto 171N or not.
$${ N }_{ V }=740+160=900N\\ { F }_{ max }=900\times 0.4=360N$$
As max value of friction can be upto 360N so our answer 171N is correct
Question 5
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A spool of mass $$M$$ and radius $$2R$$ lies on an inclined plane as shown in figure. A light thread is wound around the connecting tube of the spool and its free end carries a weight of mass $$m$$. The value of $$m$$ so that system is in equilibrium is

A
$$2M\sin { \alpha } $$

B
$$M\sin { \alpha } $$

C
$$2M\tan { \alpha } $$

D
$$M\tan { \alpha } $$

Solution

frictional force will act downwards so as to balance the torque of tension force.
For rotational equilibrium, $$\tau_o = 0 \implies f(2R) = TR$$
For translational equilibrium: $$Mg sin \alpha + f = T$$
Thus $$Mg sin \alpha + \dfrac{T}{2} = T \implies Mg sin \alpha = \dfrac{T}{2}$$
Also, $$T= mg$$
$$Mg sin \alpha = \dfrac{mg}{2}$$
$$\implies m = 2M sin \alpha$$
Question 6
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A particle P of mass m attached to vertical axis by two strings AP and BP of length l each. The separation AB=l. The point p rotates around the axis with an angular velocity $$\omega $$. the tension in two strings are $$ { T }_{ 1 }$$ and $$ { T }_{ 2 } $$
taut only if $$ \omega >\sqrt { \frac { 2g }{ l } } $$

A
$$ { T }_{ 1 }$$= $$ { T }_{ 2 }$$

B
$$ { T }_{ 1 }$$+ $$ { T }_{ 2 }$$=$$\cfrac{2}{\sqrt3}m\omega^2l $$

C
$$ { T }_{ 1 }$$- $$ { T }_{ 2 }$$=2mg

D
BP will remain

Solution

From figure in equilibrium, $$T_1\cos60=mg+T_2\cos60$$ here $$\theta=60$$ for equilateral triangle.
or $$T_1(1/2)=mg+T_2(1/2) \Rightarrow T_1-T_2=2mg$$
also, $$T_1\sin60+T_2\sin60=m\omega^2l$$
or $$T_1+T_2=\frac{2}{\sqrt 3} m\omega^2l$$
Question 7
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The spool shown in the figure is placed on a rough horizontal surface and has inner radius r and outer radius R. The angle $$\theta$$ between the applied force and the horizontal can be varied. The critical angle $$(\theta)$$ for which the spool does not roll and remains stationary is given by :

A
$$\theta=cos^{-1}\left (\cfrac {r}{R}\right )$$

B
$$\theta=cos^{-1}\left (\cfrac {2r}{R}\right )$$

C
$$\theta=cos^{-1}\sqrt {\cfrac {r}{R}}$$

D
$$\theta=sin^{-1}\left (\cfrac {r}{R}\right )$$

Solution

If the spool is not to translate,
$$F cos \theta =f$$ ....(i)
If the spool is not to rotate,
$$fR=Fr$$ .....(ii)
or $$\frac {fR}{r}cos\theta=f$$
or $$cos\theta=\frac {r}{R}\Rightarrow \theta=cos^{-1}\left (\frac {r}{R}\right )$$
Also the line of action of F should pass through the bottom most point.
Question 8
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A particle of mass $$m$$ is suspended by a string of length $$l$$ from a fixed rigid support. A sufficient horizontal velocity $$\displaystyle v_{0}= \sqrt{3gl}$$ is imparted to it suddenly. Calculate the angle made by the string with the vertical when the acceleration of the particle is inclined to the string by $$\displaystyle 45^{\circ}.$$

A
$$\displaystyle \theta = \frac{\pi }{2}$$

B
$$\displaystyle \theta = \frac{\pi }{3}$$

C
$$\displaystyle \theta = \frac{\pi }{4}$$

D
$$\displaystyle \theta = \pi $$

Solution

Given : $$u = \sqrt{3gl}$$
Let the velocity of the bob be $$v$$ when it reaches the point C.
From figure, $$AB = l - lcos\theta = l(1-cos\theta)$$
Using work-energy theorem : $$W = \Delta K.E$$
$$\therefore$$ $$-mg l (1- cos\theta) = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2$$

OR $$-2 gl (1 - cos\theta) =v^2 - (3gl)$$ $$\implies v^2 = g l+ 2g lcos\theta$$

Using $$a_r =\dfrac{v^2}{l} = g + 2g cos\theta $$
Also tangential acceleration $$a_t =g sin\theta$$
As the acceleration makes an angle $$45^o$$ with the string, thus $$tan45^o = \dfrac{a_t}{a_r}$$
$$\implies$$ $$a_r = a_t$$
$$\therefore$$ $$g + 2g cos\theta = g sin\theta$$
$$\implies$$ $$sin\theta - 2cos\theta -1 = 0$$ where $$\theta =\dfrac{\pi}{2}$$ satisfies the equation.
Question 9
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A ball attached to one end of a string swings in a vertical plane such that its acceleration at point A (extreme position) is equal to its acceleration at point B (mean position). The angle $$\displaystyle \theta $$ is

A
$$\displaystyle \cos ^{-1}\left ( \frac{2}{5} \right )$$

B
$$\displaystyle \cos ^{-1}\left ( \frac{4}{5} \right )$$

C
$$\displaystyle \cos ^{-1}\left ( \frac{3}{5} \right )$$

D
None of these

Solution

Let acceleration at both points be 'a'.
At point A, $$mg\sin { \theta } = ma$$
At point B, $$\dfrac { m{ v }^{ 2 } }{ r } = ma$$
By energy conservation, $$\dfrac { m{ v }^{ 2 } }{ 2 } = mgr\times (1-\cos { \theta } )$$
Solving, to get $$\cos { \theta } = -1 \ or \dfrac { 3 }{ 5 }$$
=> θ = $$\cos ^{ -1 }{ \dfrac { 3 }{ 5 } } $$
Question 10
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Directions For Questions

A small sphere B of mass $$m$$ is released from rest in the position shown and swings freely in a vertical plane, first about O and then about the peg A after the cord comes in contact with the peg. Determine the tension in the cord:

...view full instructions

just after it comes in contact with the peg.

A
$$\displaystyle \frac{mg}{2}$$

B
$$\displaystyle mg$$

C
$$\displaystyle \frac{3mg}{2}$$

D
$$\displaystyle \frac{5mg}{2}$$

Solution

Given : $$r = 0.8$$ m $$u = 0$$ m/s $$OA =0.4$$ m
$$\therefore$$ $$AM = r' =0.8 - 0.4 =0.4$$ m
After coming in contact with the peg, the sphere rotates in a circle of radius $$r' =0.4$$ m
Let the velocity of the sphere when it reaches the point v be $$v$$.
From figure, $$PC = r sin 30^o =0.4$$ m

Work-energy theorem : $$W = \Delta K.E$$
$$\therefore$$ $$mg (PC) =\dfrac{1}{2} mv^2 - \dfrac{1}{2}mu^2 $$

OR $$mg \times 0.4 = \dfrac{mv^2}{2} - 0$$
$$\implies mv^2 =0.8mg$$

Using: $$ ma_r = \dfrac{mv^2}{r'} = T' - mgsin30^o$$

$$\therefore$$ $$\dfrac{0.8mg}{0.4} = T' - \dfrac{mg}{2}$$
$$\implies T' = \dfrac{5mg}{2}$$